## Chapter 35Solutions for tenth week’s assignments

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### 35.1 Exercises 6.4

Problem 15
The sum is 0.\overline{8}, which is rational. So the sum of two irrationals may be rational. That should not be surprising; 2 -\sqrt{2} and \sqrt{2} are irrational, but their sum is the rational 2.
Problem 16
The sum is 0.262662666\mathop{\mathop{…}}, which is not repeating or terminating and thus is irrational. So the sum of two irrationals, like \sqrt{2} and \sqrt{2}, can be irrational, like 2\sqrt{2}.
Problem 25
\sqrt{50} = \sqrt{2 ⋅ {5}^{2}} = {(2 ⋅ {5}^{2})}^{{1\over 2} } = 5\sqrt{2} ≈ 7.07.
Problem 26
\sqrt{32} = \sqrt{{2}^{5}} = {2}^{2}\sqrt{2} ≈ 5.66
Problem 27
\sqrt{75} = \sqrt{3 ⋅ {5}^{2}} = 5\sqrt{3} ≈ 8.66
Problem 28
\sqrt{150} = \sqrt{2 ⋅ 3 ⋅ {5}^{2}} = 5\sqrt{6} ≈ 12.25.
Problem 41
3\sqrt{18} + \sqrt{2} = 3 ⋅ {(2 ⋅ {3}^{2})}^{{1\over 2} } + {2}^{{1\over 2} } = 9 ⋅ {2}^{{1\over 2} } + {2}^{{1\over 2} } = 10 ⋅ {2}^{{1\over 2} } = 10\sqrt{2}
Problem 42
2\sqrt{48} -\sqrt{3} = 2 ⋅ {({2}^{4} ⋅ 3)}^{{1\over 2} } - {3}^{{1\over 2} } = {2}^{3} ⋅ {3}^{{1\over 2} } - {3}^{{1\over 2} } = 7\sqrt{3}
Problem 43
-\sqrt{12} + \sqrt{75} = -{({2}^{2} ⋅ 3)}^{{1\over 2} } + {(3 ⋅ {5}^{2})}^{{1\over 2} } = -2 ⋅ {3}^{{1\over 2} } + 5 ⋅ {3}^{{1\over 2} } = (5 - 2)\sqrt{3} = 3\sqrt{3}
Problem 44
2\sqrt{27} -\sqrt{300} = 2 ⋅ {({3}^{3})}^{{1\over 2} } - {(3 ⋅ 1{0}^{2})}^{{1\over 2} } = 6 ⋅ {3}^{{1\over 2} } - 10 ⋅ {3}^{{1\over 2} } = -4\sqrt{3}
Problem 49
P = 2π\sqrt{{5.1\over 32}} ≈ 2.5. Note that 32 is an approximation to gravitational acceleration.
Problem 81
Approximating:
 1.{1}^{10} 1.0{1}^{100} 1.00{1}^{1000} 1.000{1}^{10000} 1.0000{1}^{100000} 2.5937 2.7048 2.7169 2.7181 2.7183 0.95418e 0.99505e 0.99950e 0.99095e 1.00000e

The constant e ={\mathop{ lim}}_{n→∞}{(1 + {1\over n})}^{n}, and the experimental results above bear this out. The first few digits converge very quickly.

### 35.2 Exercises 6.3

Problem 75
0.75
Problem 76
0.875
Problem 79
0.\overline{27}
Problem 80
0.\overline{81}
Problem 86
0.105 = {105\over 1000} = {21\over 200}
Problem 87
0.934 = {934\over 1000} = {467\over 500}
Problem 88
0.7984 = {7984\over 10000} = {499\over 625}
Problem 95
• {1\over 3} = 0.\overline{3}
• {2\over 3} = 0.\overline{6}
• {1\over 3} + {2\over 3} = 0.\overline{3} + 0.\overline{6} = 0.\overline{9}
• As covered in class, repeating nines are equal to a one one digit over, so 0.\overline{9} = 1.
Problem 96
Here, 3 ⋅ 0.\overline{3} = 0.\overline{9} = 1.

### 35.3 Exercises 6.5

Problem 1
3.00 ⋅ 12 = 36, true
Problem 2
0.25 = {25\over 100} = {1\over 4}, true
Problem 3
Rounding 759.367 to the second place beyond the decimal should give either 759.37 or 759.36 depending on the rounding rule. false
Problem 4
With round-to-nearest (even or upwards), this is true.
Problem 5
0.50 = {50\over 100} = {1\over 2}, true

### 35.4 Rounding and floating-point

#### 35.4.1 Rounding

Round each of the following to the nearest tenth (one place after the decimal) using round to nearest even, round to zero (truncation), and round half-up:

• 86.548
• 86.554
• 86.55

 Number Round to nearest even Truncate Round half-up 86.548 86.5 86.5 86.5 86.554 86.6 86.5 86.6 86.55 86.6 86.5 86.6

#### 35.4.2 Errors in computations

Compute the following quantities with a computer or a calculator. Write what type of computer/calculator you used and the software package if it’s a computer. Compute it as shown. Do not simplify the expression before computing it, and do not re-enter the intermediate results into the calculator or computer program. Also compute the expressions that do not include 1{0}^{16} by hand exactly. There should be a difference between the exact result and the displayed result in some of these cases. Remember to work from the innermost parentheses outward.

• (\overbrace{0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1}10\text{ times}) - 1
• (\kern 1.66702pt (\overbrace{0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1}10\text{ times})\kern 1.66702pt - 1) × 1{0}^{16}, where 1{0}^{16} often is entered as 1\text{e}16. If the result overflows (signals an error) on various calculators, replace 1{0}^{16} by 1{0}^{8} in this and later portions.
• (\kern 1.66702pt (\kern 1.66702pt (2 ÷ 3)\kern 1.66702pt - 1)\kern 1.66702pt × 3)\kern 1.66702pt + 1
• (\kern 1.66702pt (\kern 1.66702pt (\kern 1.66702pt (2 ÷ 3)\kern 1.66702pt - 1)\kern 1.66702pt × 3)\kern 1.66702pt + 1)\kern 1.66702pt × 1{0}^{16}
• (\kern 1.66702pt (\kern 1.66702pt (6 ÷ 7)\kern 1.66702pt - 1)\kern 1.66702pt × 7)\kern 1.66702pt + 1
• (\kern 1.66702pt (\kern 1.66702pt (\kern 1.66702pt (6 ÷ 7)\kern 1.66702pt - 1)\kern 1.66702pt × 7)\kern 1.66702pt + 1)\kern 1.66702pt × 1{0}^{16}

The object of this first part is to demonstrate round-off error. The first to problems, adding 0.1 repeatedly, may see no error if the device calculates in decimal. The latter four parts should see some error regardless of the base used.

Using Octave on a 64-bit Intel-based machine with the “short” display format:

• - 1.1102 ⋅ 1{0}^{-16}
• - 1.1102
• 0 : Sometimes errors cancel themselves out. Not every computational error is bad.
• 0
• - 4.4409 ⋅ 1{0}^{-16}
• - 4.4409

#### 35.4.3 Extra digits

Now copy down the number displayed by the first calculation in each of the following. Re-enter it as x in the second calculation.

• 1 ÷ 3, then 1 ÷ 3 - x where x is the number displayed.
• If you have a calculator or program with π, π, then π - x where x is the number displayed.

The object here is to see that the number displayed often is not the number the computer or calculator has stored.

Using Octave on a 64-bit Intel-based machine with the “short” display format:

• 1 / 3 produces 0.33333. Then (1/3) - 0.33333 produces 3.3333e-06 or 3.3333 ⋅ 1{0}^{-6}.
• pi produces 3.1416, and pi - 3.1416 produces -7.3464e-06 or - 7.3464 ⋅ 1{0}^{-6}.